Solution: Since $\{ v_1, \dots, v_m \} $ is linearly dependent, there exist scalars $\alpha_1, \dots, \alpha _m$, not all zero, such that \[ \alpha_1 v_1 + \dots + \alpha _m v_m = 0. \] Without loss of generality, assume that $\alpha_1 \neq 0$. Since $W\neq \{ 0 \} $, choose a nonzero element of $W$, say $w$. Take $w_1 = w$ and $w_2 = \dots = w_m = 0$. Suppose there exists $T \in \mathcal{L} (V,W)$, such that $T(v_k) = w_k$ for all $k = 1,2,\dots, m$. So, we have \begin{align*} T(0) & = T( \alpha_1 v_1 + \dots + \alpha _m v_m) \\ & = \alpha_1 T(v_1) + \dots + \alpha _m T(v_m) \\ & = \alpha_1 w_1 + \dots + 0 \\ & = \alpha_1 w_1 \neq 0, \end{align*} which is a contradiction. Therefore, there does not exist such a linear map \( T \) that satisfies \( Tv_k = w_k \) for all \( k = 1, \dots, m \).